Basic kinematics

The notations used here are following by the ones from the chapter 9 of the book "Classical Dynamics of Particles and Systems, 5th Edition" by Stephen T. Thornton and Jerry B. Marion. A notation with a prime indicates in the C.M. frame. $u$ is for the velocity before the reaction, $v$ is the velocity after the reaction. $m_1$ is the projectile(beam), $m_2$ is the target.

The characteristics for an elastic scattering:

Before reaction, $E_{CM} = \frac{1}{2} m_{1} u_{1}'^2 + \frac{1}{2} m_{2} u_{2}'^2$.
If the m2 is rest in the lab frame, $E_{Lab} = \frac{1}{2}m_1 u_{1}^2$, and $E_{CM} = E_{Lab} \cdot \frac{m2}{m1+m2} $

The characteristics for an inelastic scattering:

The characteristics for the inverse kinematics:

The characteristics for a transfer reaction in inverse kinematics $m_1 \gg m_2$:

note: the length of the radius is $V_{cm}$, and $v^{'}_{2} \gt V_{cm}$.

The ratio of $v_{light}/V_{cm} \approx \sqrt{q\,f} $.
$q = 1 + Q/KE_{cm}$
$Q = Q_{gs} - E_{x}$,
$KE_{cm} = $ total kinematic energy in the C.M. frame, mostly carried by light particle.
$f$ is the mass change ratio, $f = \frac{ m_{\,light\,before} }{ m_{\,light\,after} }$, for (d,p) reaction, $f \approx 2$.

Real case:

Suppose the beam is 132Sn ( m1 ≈ 132) and the target is deuteron d (m2 ≈ 2).
the beam energy of 132Sn is 5MeV/A, that is 660 MeV in total. Their mass ratio (m1/m2 ≈ 66).
In the lab frame, beam is moving with velocity u1. The target is static, and hence u2 = 0.
Vcm = 0.98 u1, is mostly equal to u1 for inverse kinematic reaction.

After the elastic scattering, no mass changes; the recoil is 132Sn, and the ejectile is also deuteron, and there is no excitation for both particles; Q-value =0. In the lab reference frame, the recoil of 132Sn has very small deviation off the beam axis. Its max scattering angle in the lab frame (ψ) is less than 1 deg.

In the lab reference frame, scattering angle (ξ) of the ejectile of d is always less than 90 deg,
that is always at the downstream of the beam direction; the forward angle. And it holds true for all elastic reactions that the ejectile is always less than 90 deg. And hence, several particle detector have more particle identification tools placed at the forward angle. Because they want to measure the elastic cross sections, which can be used as normalization.

(Figure: calculations from CATKIN)

one more example (from Wilton N. Catford ):

pickup reactions will be forward focus.
(d,3He) = get one more proton,
(p,d) and (d,t) = get one more neutron.

stripping reactions will be backward focus
(d,p) = lose one neutron.

(Figure from
From experiment, we can measure the Energy of the particle and the lab frame angle.
We then can convert it to Q-value spectrum.
The formula can be found at the page 140 of "Nuclear Reaction Analysis" by Maryon Young
Or from this small site (here)

note for Q-value:
The Q-value can be negative,
since portion of energy is deposited to the higher excited state.
If the final products are all in the ground states, then Q value is simply the mass difference between those before and after a reaction.
$Q = (m_{target} + m_{beam})\cdot c^{2} - ( {m_{recoil} + m_{ejectile} )\cdot c^{2}}$
When its Q value is positive, then it means we we get extra energy after this reaction.
If its Q value is negative, we need to input energy it them to make this reaction happens. For example, we use an accelerator to accelerate the beam particle, such that the $E_{beam} + Q $ > 0, and the rest of energy distribute to $ E_{recoil} + E_{ejectile}$.

How about if the recoil is not in the ground state, but in its excited state?.
In this case, we need to have $E_{beam} + Q_{ground} - E_{x} $ > 0.
or we can define $Q_{E_{x}} = Q_{ground} - E_{x} $.
For example for 86Kr(p,d) 85Kr, reaction,
ground state Q value = -7.632 MeV,
the first excited state Q value = -7.632 - 0.304 = -7.936 MeV.